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The primary and secondary coil of a transformer have 50 and 1500 turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi = \phi_0 + 4t$, where $\phi$ is in webers, $t$ is time in seconds and $\phi_0$ is a constant, the output voltage across the secondary coil is
A
120 volts
B
220 volts
C
30 volts
D
90 volts
Detailed Solution
For a transformer: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$
Here $N_p$ = number of turns in the primary coil = 50 and $N_s$ = number of turns in the secondary coil = 1500.
The emf across the primary equals the rate of change of the flux linked with it: $V_p = \dfrac{d\phi}{dt} = \dfrac{d}{dt}(\phi_0 + 4t) = 4$ V
$\phi_0$ is a constant, so its derivative is zero.
$V_s = \dfrac{N_s}{N_p} \times V_p = \dfrac{1500}{50} \times 4$
$V_s = 30 \times 4 = 120$ V
So the output voltage across the secondary coil is 120 volts.
Here $N_p$ = number of turns in the primary coil = 50 and $N_s$ = number of turns in the secondary coil = 1500.
The emf across the primary equals the rate of change of the flux linked with it: $V_p = \dfrac{d\phi}{dt} = \dfrac{d}{dt}(\phi_0 + 4t) = 4$ V
$\phi_0$ is a constant, so its derivative is zero.
$V_s = \dfrac{N_s}{N_p} \times V_p = \dfrac{1500}{50} \times 4$
$V_s = 30 \times 4 = 120$ V
So the output voltage across the secondary coil is 120 volts.